15.1 Double integrals EX 15, 19, 21
15.2 Iterated integration in Cartesian coordinates EX 3, 5, 15, 23
15.3 Generalized integrals and the mean value theorem EX 1, 3, 13, 27
15.4 Double integrals in polar coordinates EX 5, 9, 15, 19, 21
15.5 Triple integrals EX 5, 7
15.6 Change of variables in triple integrals EX 7
15.7 Applications of multiple integrals Example 4, EX 5, 13
Dr. Wanmin Liu
Multiple integrals
SF1686 HT21-1
wanminliu@gmail.com
Multiple
Integration
Now
;
multivariable
integration
.
Review
:
One
variable
integration
SS
f-
ix.
g)
DA
=
fs
fix
.y
>
dxdy
R
q
r
g
b
different
notations
for
Acxsdx
=
fabdx
(
Aix
)
)
KEY
information
:
9
9
①
Region
R
of
the
domain
of
two
variables
Two
different
notations
,
containing
exact
②
function
f-
(
x
,
y
)
of
two
variables
the
same
information
;
The
region
R
and
fourteen
f-
ix.
Y
)
play
①
✗
is
a
variable
,
from
a
to
b
the
same
important
role
for
the
integration
.
②
the
integr and
(
the
fumtion
been
integrated
)
How
to
compute
?
is
the
function
Aix
of
✗
-
iteration
of
two
one
variable
integr ations
Remark
①
b
is
not
necessary
to
be
greater
than
a
gabdxfh
fix
,y
,
dy
gun
since
fab
Aix
,
dx
=
-
f
?
Aix
,
dx
We
think
FIRST
that
×
is
The
whole
expression
a
constant
and
y
is
a
variable
.
②
The
whole
expression
fab
Aixdx
is
afumttonotx
moreo v er
,
y
varies
from
the
Is
nothing
to
do
with
the
notation
✗
with
nothing
to
do
bound
gix
to
the
bound
hix
)
.
(
Here
gun
is
not
necessary
to
be
so
we
can
use
different
notation
for
the
with
Y
'
smaller
than
nix
,
)
.
Variable
Jab
Aix
>
dx
=
Jab
Act
>
dt
finally
,
we
compute
we
compute
the
one
-
variable
integral
the
whole
one
variable
for
Y
,
the
solution
is
an
expression
with
"
constant
✗
"
.
Integral
for
✗
.
wanminliu@gmail.com
§
15.1
Double
integrals
50-1
fix
.
-1
)
=
✗
+
y
#
1g
evaluate
the
given
integral
BY
gino
the
domain
①
T
is
symmetric
inspection
.
with
respect
to
-06
origin
10.0
)
.
If
1×+41
DA
and
ftx
.
-
Y
)
=
1-
X
)
+
C-
y
)
where
T
=
parallelogram
with
=
-
(
✗
c-
Y
)
four
vertices
I
=
-
f-
ix.
y
)
A-
(
2,2
)
he
②
fix
>
y
)
is
also
symmetric
with
÷!×
AB
"
"
ADHBC
respect
1-
the
origin
,
so
①
②
ÉBu
,
-1
)
cc
-2
,
-4
§
fix
.ydA
=
o
☐
wanminliu@gmail.com
•
what
is
the
demah
4-
×%×
So
✗
C-
@
it
]
I
=
2
In
2
☐
•
4-
we
fix
✗
as
'm
the
domah
what
is
the
domain
otyc.co
,
x
]
T.tt#iaI+D--IidxYif-de-ixJidlnlxx?lnlxtys/I--fn(
Ix
)
-
lnx
=
fu
2
=
filmy
dx
=
ln
211-0
)
=
bn
2
.
wanminliu@gmail.com
#
19
f-
f)
(
a-
xZÑ
)
DA
22=11
-
×ÑdA
xkyka
'
Kaisa
'
☒
1=11
a
DA
-
ffxIyTdA
×¥r
'
'
"
✗
Eisa
'
{
*
noo
HdA=f"d%rdr
=
#
×¥aÑ
-x
=
-
r
y=rsTnO
+
22
I
,=
a
HdA=a
.
Areal
disc
)=
a.
Total
'
Then
I.
=
f
?
"
do
f.
"
rdr .fr
)
xIy¥a
'
=
a.
a.
a
'
rad ius
__
lal
Warning
:
a
could
be
=
-
/
Fto
t.at
Fdr
anegatwe
number
!
¥
¥dr
}
Domain
of
integral
^
"
disc
of
radius
/
at
lap
-0
)
=
-2×1
-1,1N
'
)
#-
✗
So
1=2,1-32
=
+
a
'
-
¥
lap
=
{
a
'
azo
¥
a
}
a
co
☐
wanminliu@gmail.com
#
21
2¥
(
1-
✗
-
y
)
da
that
f)
xD
A
=
If
y
da
.
T
T
where
T
=
triangle
OAB
because
the
symmetry
property
of
A
=
11.0
)
B
(
0.1
)
,
,
-
✗
=y
T
with
respect
to
the
line
✗
=y
.
50-1
.
M
B
I
•
Aveet
)
=
§
.
1
.
I
=
I
①
,%É;×
Ss
✗
dA=jid×
5×+141×1
T
°
=
So
'dx$-
✗
1
"
"a
,
'
µ
o
=
✗
1-
✗
+1
)
I
=
¥1
DA
-
SI
✗
DA
-
5)
ydA
"
I
!÷•¥¥
"
"
"
"
"
=
-
isiaxs-i.is
:*
.
T
=
-
-1 , 4-01+-2111-01
from
0
=-D
②
=
Areal
-1
)
-
2¥
✗
HA
⑧
Tox
✗
goes
from
o
to
p
where
we
use
the
observation
if
we
fix
a
✗
c-
[
on
]
,
then
Y
c-
[
0
>
1-
×
]
so
I
②
I
-
2x
f-
=
I
☐
wanminliu@gmail.com
§
15.2
Iteration
of
Double
integrals
in
Cartesian
coordinates
=L
-
y
f
?
d
@
x
)
=
-4
)
co
>
✗
I
?
=
Cy
/
cost
-
bro
)
⑧
#
3
wa=
-1
cosy
dy
DX
bro
=
I
f
?
S
-
Ix
W
dying
)=wsYdy
so
=
If
dfmy))d×~
⑧
=
(-4/-1-1)=4
-
☐
5m19
)
/
¥
=
sin
✗
-
sintx
)
=
J
?
25m
✗
dx
=
25in
✗
day
=
lwsxjdx
=
-4.2
)ff5lnxdx_)
=
-
sinxdx
wanminliu@gmail.com
#
5
evaluate
the
double
integrals
by
so
I
=)
?
dx
Ji
dy
1×1-5
)
iteration
=
fi
dx
(
f
?
Kristy
)
f-
f)
(
✗
7-
5)
DA
-
R
we
think
✗
as
a
constant
R
=
rectangle
{
ix.
y
,
/
0£
✗
£
a
}
only
y
as
variable
0-9
Eb
✗
if
?
dy
+
Job
ily
Lol
.
By
the
condition
,
a
30
&
b
so
=
✗
2.
b
+
f
b
'
b%×
=Ii*b+¥)d×
=
b)
I
vidx
+
§
flax
¥
a-
①
What
is
the
domain
of
✗
?
✗
E
Co
,
a
]
=
b.
f-
as
+
§
a
③
If
we
fix
such
a
✗
c-
[
0
,
a
]
,
where
is
b
?
=
tabla
4-
5)
.
Y
C-
[
o
,
b
]
☐
wanminliu@gmail.com
#
15
tidy
ly
'
e-
"
dx
=L
so
I
=
If
e-
"
body
T
SI
(
what
is
the
resin
G-
integral
)
1-
=
triangle
OAB
•
Y
f
[
0.1
]
Idly
;
the
original
integrator
is
NOT
easy
to
compute
.
•
G-
we
fix
such
a
y
,
then
so
we
recover
it
by
the
double
integral
✗
C-
[
Y
.
I
]
.
&
Using
another
expression
for
such
double
so
we
draw
the
✗
=Y
line
integral
for
the
T
resin
✗
Y
•
✗
C-
fo
,
I
]
•
4-
we
fix
such
✗
,
then
Y
C-
[
0.x
]
"
*¥×÷
×
✗
Y
•
y
is
fr om
A
"
'
"
'
•
*
✗
É÷÷""×
✗
•
A
-11.0
)
✗
E
lo
,
I
]
wanminliu@gmail.com
=
I
side
-
t
)
*
2=1
!
dxfidy
é
"
q
-
f
,
deft
/
=
étdtt
,
=
f.
'
e-
¥114
Up
=
-
e-tdt-z-ue.tl
!
✗
=
f
'
e-
"
e.
=
t.eu/e-'-e7xdx--.Idx--tHqe-*dx
'
=
-
t.lt
-
1)
1-
=
×
?
So
ftp.t
]
this
is
the
range
g-
✗
=
£
(
1-
t
)
=
±
He
-
+
dt
=L
☐
at
this
is
the
range
of
t
wanminliu@gmail.com
#
23
Find
the
volume
otttenesi-nflldxdydz-fijxf.by/
¥
da
V
-
under
the
surface
z=¥y
¥
&
above
the
✗
y
-
plane
""""""""
"
"
"
"
"
If
d¥
dlxtykdy
✗
Y
y=✗
Sine
we
1
'
are
thinking
✗
Lid
lnlxty
,
as
a
const
.
"
lnlxxy
,
/
ii.
e¥Y¥n
az
☐
we
observe
2-
=¥y
so
=
lnlxtx
-
lnlxto
)
=
ln
2X
-
ln
✗
for
IX.
7)
ED
=
ln
2
thnx
-
lnx
=
ln
2
÷%¥¥?E¥¥
=
)
?
ten
2)
DX
=
ln
2.
(
2-
1)
=ln
2
☐
For
an
integral
:
the
region
&
the
function
play
the
same
important
role
wanminliu@gmail.com
§
15.3
Improper
integrals
15
de
g
?
ay
1
e-
*
✗
)
Determine
whether
the
given
integral
converges
or
diverges
.
Try
to
evaluate
then
,
this
value
would
be
the
value
of
the
original
integral
,
that
converges
.
A-
IT
d×µ¥
!
.
e-
"
"
"
duty
,
#
I
5-
f)
e-
*
✗
da
a
weÉÉt
constant
.
and
Y
as
a
variable
=
first
quadrant
of
the
XY
-
place
dcxty
)
=dy
-1×+41
-
LA
d
e-
"
+
"
(
"
"
Yin
,
f-
IX.
y
)
=
e-
✗
-
y
=
e-
1×+41
god
we
don't
know
a
prior
whether
the
-
e-
(
✗
+
A)
+
g-
(
✗
to
,
"
"
dy
a.
µ
µ
.my
,µ,µ
guy
,
www..ge
,
Mn
1in
1-
e-
"
"
"
+
E)
=
e-
✗
A-
→
is
•
If
we
can
evaluate
the
following
=
J
?
e-
✗
dx
=
1-
1)
¥
:
tide
-
✗
Integral
=
C-
D.
(
Has
e-
A
-
E)
=
a
.
a-
So
I
converse
os
to
1
.
☐
wanminliu@gmail.com
#
3
I=§
µ÷dA
=
y
/
laing
,
ar ctan
A
-
limantanta
)
)
where
S
=
the
strip
0<74
in
the
XY
-
plane
a-
→
•
Yan
,
it
i
I
>
×
0
-
¥
_>§×
Sol
.
We
compute
folds
f%¥×d×
If
5Wh
integral
exists
,
then
it
gives
original
=
Y
l
¥
-
1-
¥1
)
=
ay
y
is
a
constant
integral
.
fol
dy
f
?•¥×dx
£
JI
1¥
,
dx
=
Y
l
]
¥
d×
=
folky
dy
=
g-
xfo
'
dy
-
=
-5
.
=
y
FI
d
Canton
×
)
so
I
=
Iz
=
y
lim
ar ct anx
/
A
☐
A-
→
is
-
A
wanminliu@gmail.com
#
13
Evaluate
3=5, 5
¥
,
DA
Sol
Ñ-
not
a
smart
method
.
where
S
is
the
square
04<-1
,
04<-1
.
(b)
.
let
W
be
the
triangle
0
Bc
(a)
by
dire ct
iteration
of
double
integral
(b)
by
using
the
symmetry
of
the
integrand
§
¥ydtt=
µ
¥
,
dat
¥
DA
and
the
domain
to
write
Now
fix
.
D=
¥y
is
symmetric
with
1=2
4-
¥¥
respect
to
the
the
✗
=p
.
Where
T
is
the
triangle
with
vertices
010 , 01
A
11 . 0
)
1311
.
1)
.
that
"
f-
(
Y
'
×
)
=
¥
=
¥y
,
fix
,
y
)
and
W
is
the
symmetric
region
of
T
Ya
with
respect
to
the
line
✗
=y
¥
☐
"
'
"
f)
fly
,x)dA
¥dA-
II
fix
.y
)
DA
T
0
All
.no#vqr
"
4-
%
call
my
so
I
=
2
§¥÷
f)
ftp.xdA
1-
Hvar
.mx
82nd
var
or
y
T
¥µy
)
How
could
I
write
"tÉ?"w"=
-1
wanminliu@gmail.com
¥
DA
=
f)
dx
f)
¥
,
dy
so
I
=
2
If
¥
,
DA
=
fol
dx
If
É
dcxty
)
=
2hr2
9
☐
For
the
integ ral
of
variable
y
,
X
is
a
constant
so
duty
)
=
dy
=/
☐
'
dx
ln
*
→
4
?
=
Ii
Cmx
)
dx
lnztlnx
-
lnx
=
bn
2
=
Jd
lnz
dx
=
lnz
wanminliu@gmail.com
#
27
Does
f-
ix.
y
)
=
✗
y
have
an
average
let
us
compute
I
,
by
the
following
.
value
over
the
region
fjsdx
fo¥
'
✗
y
dy
R={
IX.
YJEIRYOE
✗
<
is
,
of
Y
s
¥
.
}
?
=
f
?
xdx
.
±
f.
¥
.
dy
'
if
so
,
what
is
it
?
=
fix
.
I
.
1¥
)
'
dx
Sol
.
•
?
=
-1
-
I
15
¥gp
dx
'
+
=
×
'
I
•,#
¥2
=
-41
'
I
?
¥4T
It
Itt
=
w
c-
[
i.
a)
•
>
✗
dt
=
dw
=
I
f
?
I.
dw
-4--1,41
?
dcw
"
)
The
average
of
the
function
f-
ix.
y
)
=
-414-4/7=1--4 %0
+
É
'T
=
-5
in
the
region
R
is
defined
to
be
2¥
So
I
,
=
-14
.
I
,
=
If
5-
ix.
Y
)
DA
we
compute
22
by
the
foll owing
R
Sid
✗
so¥
'
1
dy
=
Sf
¥×
,
dx
=
food
Can-Can
x
)
12
=
Area
(
R
)
=
IS
9-
DA
R
=
digs
automat
=
¥
.
So
average
=
2/2
,
=
-41¥
=
¥
☐
wanminliu@gmail.com
§
15.4
Double
integrals
in
Then
Polar
Coordinates
dA=
dxdy
=
}¥
dudu
✗
=
rcoso
r
'
=
✗
4-72
where
Y
=
rsTn
0
tan
0=4
/
✗
214.7
)
I
r
?
is
the
Jacobian
determinant
,
i.
e.
1
Y
'
•
!
=/
*
¥1
✗
⇐
÷
Recall
dxdy
=
DA
=
r
drdo
¥,=¥÷
.
'
More
general
,
assume
that
✗
=
✗
(
U
.
V
)
with
?¥¥=|¥×
¥
'
/
.
y
=
y
tu
,
V1
¥
¥
✗
y
-
plane
uv
-
plane
&
onto
wanminliu@gmail.com
#
5
Let
D
be
the
disk
To
compute
fix
wio
do
{
IX.
y
)
/
✗
4-
y
's
a
'
,
where
a
>
o
}
.
We
WANT
To
decrease
the
degr ee
by
increasing
Evaluate
the
angle
I
=
IS
X2
DA
.
COS Y
=
f-
Cos
20-+1
)
D
Then
Sol
.
✗
=
ruse
I
=
-9<4
.
-1/0*6520-+1
)
do
dA=
rdrdo
domain
of
0
:
co
.
4)
=
-4
?
I
Jo
"
wszodlzo
)
-1
If
"do]
domain
of
r
:
[
o
,
a
]
=
-4
"
.
Ifi
sin
-
of
"
+
za
]
I
=
f.
"
do
fir
dr
T
27
=
¥
a
"
=
Jo
do
dosa
)
'
II
rsdr
☐
¥
=
qa
"
si
"
wio
do
wanminliu@gmail.com
#
9
Define
Q=
/
ix.
y
)
/
✗
7.
Yao
.
✗
Eisa
'
}
with
a
>
o
Let
t
=P
then
t
C-
[
0
,
a
'
]
.
Evaluate
[
=
f.ge#5dA
So
Q
Sol
.
I
=
¥
fiietdt
it
=
¥
fiidletj
-
x
=
lead
-
é
)
we
use
polar
coor dinate
.
=
-5
led
-11
.
Now
D-
C-
[
0
,
I
]
and
r
C-
[
o
,
a
]
.
☐
I
=
do-sierrdr-I-if.ae
"dr
'
.
wanminliu@gmail.com
#
15
Find
the
average
distance
3
,
=
fo
"
do
/
oar
rdr
from
the
origin
to
points
in
the
disk
D
=
22
.
Jo
"
-31
dr
}
✗
4yd
E
a
'
with
a
>
0
.
a
}
So
/
.
=
22
.
-3
✗
Y
'
°
average
=
a
}
>
✗
*
a
,
=
f-
a
☐
The
average
=
¥-
with
2
,
=
If
r
DA
D
Iz
=
Area
(D)
wanminliu@gmail.com
#
19
Evaluate
1=55
✗
y
DA
D
I
=
-14
10¥
sin
20
dczo
)
.
where
D=
{
IX.
y
)
/
✗
70,0
EYE
✗
}
=
iz
f-
cos
20
)
/
¥
.
-9, 4
✗
Ty
's
a-
,
a
>
0
0
Sol
.
^
"
=
If
+
↳
01¥
y=×
0
9-
to
=
9¥
.
a
>
✗
☐
I
=
1¥
do
I
?
Cros
lrsino
)
rdr
a
r
}
dr
=
0¥12
show
>
Old
"
toy
wanminliu@gmail.com
✗
£
2-
=r2
fr
>
o
)
#
21
Find
the
volume
lying
between
Ttx
r
c-
(
0.1
)
the
paraboloids
then
2-
goes
from
:¥z=4
-
rz
(
r
>
,
r
'
to
14¥
o
:#
r
2-
=
✗
4-
y
'
and
32
=
4
-
(
✗
'
+
y
'
)
.
We
use
cylinder
coordinate
Sol
.
Vol
=
fidofordr
/
"
devol
=
SSS
dxdydt
rz
Region
V
=
2x
fir
/
-34
-
I
'
-
rydr
where
region
V
is
the
rotation
of
=
zafifir
-
Ir
'
/
dr
the
shaded
area
in
the
r
-
2-
plane
=
27
.
-34
(
I
-
-14
)
along
the
2-
-
axis
27
angle
=
¥
☐
wanminliu@gmail.com
§
15.5
Triple
integrals
=
So
'
dx
f)
(
✗
'
c-
y
'
)
dy
=
fi
dx
[
×
'
tidy
+
fiy
-
dy
]
#
5
Let
R
be
the
cube
T
F
R
=
{
ix.
7
.
2-
)
/
Of
✗
,
Y
.
2-
El
}
.
=
f
!
1×2+-5
)
dx
Evaluate
=
-5+-5
I
=
(
✗
'
+
Y
'
)
dv
.
=
-32
Sol
.
☐
I
=
f)
dx
f)
dy
f.
'
dz
1×7--14
q
Observation
:
the
integr and
is
Independent
of
2-
=
Kidz
dx
fig
cxiyy
)
I
wanminliu@gmail.com
"
Z
#
7
Define
we
observe
that
-
'
=L
fssxydv
=
0
R=
/
IX.
42-71
Of
2-
f
1-1×1- 171
}
.
4
Evaluate
since
the
region
V
is
☐
.
,¥%
>
y
1=1111×7+2-2
)
dv
.
symmetric
for
✗
oey
.
R
X
f
By
the
symmetry
of
the
region
R
Sol
.
I
=
SLS
z
-
dy
=
4
Sss
z
-
dy
Ro
we
first
draw
the
2-
=o
plane
.ie
wher e
R
,
is
the
subr egion
OEAB
the
Xy
-
plane
in
XYZ
-
space
.
¢
For
a
fixed
☒
in
with
✗
so
&
Y
70
Then
ay
1-
✗
-
Y
2-
goes
from
o
to
1-
X-Y
3=4
So
'dx
dyf
.
2-
Zdz
✗
y
,
1
.
B
✗
y
>
o
For
a
fixe d
✗
c-
10 .1
)
write
•
1-
✗
Y
C-
10
,
l
-
X
)
w
=
l
-
X
-
Y
-
i
1
A
}
=
4
.
-13
.
sides
!
"
4-
×
-
Y
)
}
dy
c
'
'
O
'
,
>
✗
dw
=
-
dy
since
-
here
✗
is
a
gµ
✗
Y
>
o
xyco
-
Ji
-
✗
(
1-
X
-
Y
)
}
d
l
1-
✗
-
Y
)
where
is
w
-1
-
D
Tgif
=
-
ziti
-
✗
-1-1×5
=
4
.
-31
.
-4
So
'××d×
+
ziti
-
x-D
"
=
-14×4
=
-;.÷=¥=w¥ ¥:l☐s;wsedw
,
wanminliu@gmail.com
let
✗
=
au
y=
bv
2-
=
cw
§
15.6
Change
of
variables
with
a
,
b.
c
so
in
triple
integrals
=/
:
:
:|
-
-
an
#
7
Let
the
region
R
be
defined
below
.
Now
the
line
y=×
becomes
V=
g-
u
.
In
the
first
octant
,
between
the
Vol
=
Sss
dy
=
abc
#
dudv dw
planes
y=o
and
y=×
,
and
inside
R
w
~
the
ellipsoid
!
p
v=§u
R
+
¥
+
¥
=/
.
(
a.
b.
C
>
°
)
¥÷¥¥¥¥
.
Find
the
volume
of
R
.
tano-gp v-f.ci
Vol
-_
Sssdxdydz
Sol
.
R
"
£
We
change
µ
,
V
,
W
)
-
coor dinate
to
spher e
word
-
lr.0.IO
)
e
R
Y=✗
With
of
Rsl
,
050
E
are
-6m¥
,
¢
c-
[
0
>
¥
]
Vol
=
abcsssdudvdw
=
abc
f¥d§%d
)
;
pisinodr
.
.
=aփf
@
retail
)%¥dR_
=
antanas
.
✗
y=✗
dudvdw
-_
pisinodrdodo
☐
wanminliu@gmail.com
G.
¥0,271
.
Sol
§
15.7
.
Applications
µ
t
dv=
R'
sinodrdodo
¥÷
"
"µa
)
I
=
Sss
iipdv
B
R
E
lo
>
a
]
.
$
C-
[
0,2
]
,
D-
c-
10 . 2-a
]
EXAMPLE
4
7880
<÷F
10.4
D=
r
=
x4=R
Sino
the
acceleration
of
a
rolling
ball
)
✗
=P
/
F'
do
/
idol
foadrlpisiio
-
Pisano
)
'
Let
B
be
a
solid
ballot
radius
a
la
>
→
w
1¥
and
costant
density
p
.
"
÷
;÷¥÷%= →
:*
.
.
191
.
Find
the
moment
of
inertia
-
hÉ
silt
:b
at
and
radius
of
gyration
of
B
|
=/
Eloise
-1
)dKsD=
Hid
@
son
'
-
Iidwsoi
"
"
"
"
"
(b)
.
With
what
linear
acceleration
will
the
ball
/
+
=
cosy
what
is
the
range
of
-1=-5/(-1,3-13)
-
(
y
-
y
=
2-
É
roll
(
without
slipping
)
down
a
plane
inclined
at
1*1%7-1
=
-
-5+2=-9
angle
✗
to
the
horizontal
?
as
=P
.
27
.
-34
-
5-
=
pas
.
I
=
Sss
rip
dv
is
called
moment
of
inertia
m=§zpa
'
R
4
the
particle
is
mo ving
in
a
circle
of
radius
D
I
=
Imf
=
a
.
D-
=
Ifm
Is
called
the
radius
of
gyration
(b)
.
Textbook
.
Wien
m=
fsspdv
.
R
wanminliu@gmail.com
https://wanminliu.github.io/doc/DG/DG.html
#
5
(
surface
area
)
Now
taking
partial
derivativ e
with
respect
to
✗
for
the
equation
3z2=
✗
4-
y
'
,
we
hav e
Find
the
surface
area
of
the
conical
surface
322
=
✗
4-
y
'
with
0<-2-12
.
62
2-
✗
=
2X
.
Similarly
,
Sol
.
If
the
surface
is
giv en
by
2=2-1
✗
in
6
2-
2-
y
=
zy
.
1+2×42-5=4
⇐
E)
'
+
¥5
Then
surface
area
S=ffH⇐ÉdA
D
=
it
-1g
¥-5
=
it
-1g
.
=
-5
In
general
,
for
a
surface
defined
by
a
parameter
equation
Ñ=
flu
,
v
,
then
Moreov er
,
for
the
range
os
2-
ez
,
we
hav e
of
✗
4-
y
'
E
12
surface
area
=
If
Ec
dudv
=
ffds
q
cubed
swfml
g
=
If
tzE+zÑ
did
>
For
more
details
,
you
are
welcome
to
read
my
oexh-iyk.IT
online
note
:
=
¥
ff
dxdy
(
Lecture
8)
.
°£×¥Éf
a
circle
with
radius
if
¥É¥*
:
=
Fg
a
.
12
=
8532
'
☐
wanminliu@gmail.com
#
13
Find
the
mass
of
a
spherical
planet
we
use
spherical
coordinate
radius
a
whose
density
at
distant
R
from
m=
f
?
do
ffdpfoapis.in/opdR
center
is
p=☐÷
with
A.
B
>
°
.
=
f
?
do
ftp.ino/1dofIAp.R+ p-.dR
We
denote
the
spherical
Sol
'
planet
by
Ball
¥
=
TOUCAN
=
47A
fo
"
B+R÷
dr
iwktheothe
case
"
Boo
"
}
•
pi
"
-
#
=
T.IE?---i-yI-y,.-.+.-dt
'
.
R
a
let
5
=
1¥
efo
,
¥
]
we
use
the
fait
13>0
.
ds=
trzdr
a
=
47A
HB
(
1-
Bids
=
42A
(
a
-
rBfo¥
dfarctans
)
)
M
=
fff
p
dy
=
4k
Ala
-
Es
.
arctanarz
)
Ball
☐
